Scaling Augmented Matrices: Is My Thinking Correct?

In summary, scaling an augmented matrix by a number is equivalent to multiplying each row by that number, resulting in multiples of the same equations and solutions. However, if the number is zero, all conditions are lost. This concept can also be applied to matrix equations, where multiplying the coefficient matrix by a constant does not change the solution set.
  • #1
jeremy222
5
0
An augmented matrix scaled by a number also means the solutions set is scaled by that same number. I believe this is true due to it basically being the same as elementary row operations preformed on each row. Unless it is a zero scalar in which case you lose all conditions. Is my method of thinking about this correct? Its not homework but I think it would help me understand a concept.
 
Physics news on Phys.org
  • #2
No, that's not true. If you have an augmented matrix, say
[tex]\begin{bmatrix}a & b & c & j \\ d & e & f & k \\ g & h & i & l\end{bmatrix}[/tex]
That is equivalent to the three equations ax+ by+ cz= j, dx+ ey+ fz= k, gx+ hy+ iz= l.
It is also equivalent to the matrix equation
[tex]\begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}j \\ k \\ l\end{bmatrix}[/tex]

"Scaling" the augmented matrix by a number, multiplying every row by that number, gives multiples of the same equations which have the same solutions. If you multiply the coefficient matrix in the second form by a constant, but not the right hand side, that will divide the soluutions by the number.
 
Last edited by a moderator:
  • #3
one can look at the 1x1 case:

[a][x] =

if we multiply by 2, let's say, we get:

[2a][x] = [2b].

suppose a ≠ 0. then in either case, we find:

x = b/a, the solution to the 2nd equation is NOT x = 2(b/a).

it is exactly the same for

Ax = b, where A is a matrix, and b is a vector.

if A is invertible, x = A-1b.

and 2Ax = 2b has the same vector solution.

in other words, multiplying an equation on both sides by a number,

doesn't change the solution set for that equation.

if ax + by + cz = d, then for any number m:

m(ax + by + cz) = md, for the same x,y and z.
 

1. What is the purpose of scaling augmented matrices?

The purpose of scaling augmented matrices is to simplify and organize large sets of data, making it easier to analyze and draw conclusions from. It is a common technique used in various fields of science and mathematics.

2. How do you determine the appropriate scale for an augmented matrix?

The appropriate scale for an augmented matrix is determined by the range and magnitude of the data being represented. It should be chosen in a way that allows for easy interpretation and comparison of values.

3. Can scaling an augmented matrix affect the accuracy of the data?

Yes, scaling an augmented matrix can affect the accuracy of the data if not done properly. It is important to use a consistent scale and ensure that all data points are accurately represented.

4. What are some common techniques for scaling augmented matrices?

Some common techniques for scaling augmented matrices include using logarithmic scales, linear interpolation, and standardizing the data. The appropriate technique will depend on the type of data being represented and the goals of the analysis.

5. Is there a limit to the size of an augmented matrix that can be scaled effectively?

There is no specific limit to the size of an augmented matrix that can be scaled effectively, as it will depend on the specific data and the chosen scaling technique. However, it is important to consider the readability and interpretability of the scaled matrix when dealing with very large datasets.

Similar threads

  • Linear and Abstract Algebra
Replies
2
Views
2K
  • Precalculus Mathematics Homework Help
Replies
25
Views
982
  • Linear and Abstract Algebra
Replies
14
Views
2K
  • Precalculus Mathematics Homework Help
Replies
1
Views
723
  • Linear and Abstract Algebra
Replies
15
Views
4K
  • Linear and Abstract Algebra
Replies
4
Views
7K
  • Linear and Abstract Algebra
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Linear and Abstract Algebra
Replies
4
Views
2K
Replies
7
Views
846
Back
Top